Increasing Function Example

Question 1.

Determine where [f(x)=x^{3}-3x] is increasing.

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)=3x^{2}-3].
2. Solve [3x^{2}-3=0] ⇒ [x^{2}=1] ⇒ [x=\pm 1].
3. Sign chart for [3x^{2}-3]: choose intervals (-∞,-1), (-1,1), (1,∞).
   • At [x=-2]: [3(4)-3=9>0] ⇒ derivative positive.
   • At [x=0]: [-3<0] ⇒ derivative negative.
   • At [x=2]: [9>0] ⇒ derivative positive.
4. Conclusion: [f(x)] is increasing on [(-\infty,-1)\cup(1,\infty)] and decreasing on [(-1,1)]. (Not strictly increasing on intervals containing the stationary points; on the open intervals above it is strictly increasing because derivative > 0 there.)

Question 2.

Find intervals where [f(x)=\ln x – \dfrac{x}{2}] (domain [x>0]) is increasing.

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)][=\dfrac{1}{x}-\dfrac{1}{2}].
2. Solve [\dfrac{1}{x}-\dfrac{1}{2}=0] ⇒ [x=2].
3. Test intervals (0,2) and (2,∞):
   • At [x=1]: [1 – \dfrac{1}{2} = \dfrac{1}{2} > 0] ⇒ increasing.
   • At [x=3]: [\dfrac{1}{3} – \dfrac{1}{2} ][= -\dfrac{1}{6} < 0] ⇒ decreasing.
4. Conclusion: [f(x)] is increasing on [(0,2)] (strictly increasing there) and decreasing on [(2,\infty)].

Question 3.

Where is [f(x)=e^{-x^{2}}] increasing or decreasing?

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x) = -2x e^{-x^{2}}].
2. Solve [-2x e^{-x^{2}}=0] ⇒ [x=0] (since [e^{-x^{2}}>0]).
3. Sign of derivative determined by [-2x]:
   • For [x<0], [-2x>0] ⇒ derivative > 0 ⇒ increasing.
   • For [x>0], [-2x<0] ⇒ derivative < 0 ⇒ decreasing.
4. Conclusion: [f(x)] is increasing on [(-\infty,0)] and decreasing on [(0,\infty)] (strictly on each open interval).

Question 4.

Determine where [f(x)=\dfrac{x^{2}}{1+x^{2}}] is increasing.

Step-by-step Solution:

1. Use quotient rule:
   [\dfrac{d}{dx}f(x)][=\dfrac{2x(1+x^{2}) – x^{2}(2x)}{(1+x^{2})^{2}} ][= \dfrac{2x}{(1+x^{2})^{2}}].
2. Solve [\dfrac{2x}{(1+x^{2})^{2}}=0] ⇒ [x=0].
3. Sign: numerator [2x] gives sign.
   • For [x<0], derivative < 0 ⇒ decreasing.
   • For [x>0], derivative > 0 ⇒ increasing.
4. Conclusion: [f(x)] is increasing on [(0,\infty)] and decreasing on [(-\infty,0)]. (At [x=0] derivative = 0; function has a minimum there.)

Question 5.

Find where [f(x)=x^{4}-4x^{2}] is increasing.

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)][=4x^{3}-8x ][= 4x(x^{2}-2)].
2. Critical points: [x=0,\ \pm\sqrt{2}].
3. Sign chart intervals: [(-∞,-√2)], [(-√2,0)], [(0,√2)], [(√2,∞)]. Test:
   • x=-2: derivative negative.
   • x=-1: derivative positive.
   • x=1: derivative negative.
   • x=2: derivative positive.
4. Conclusion: [f(x)] is increasing on [(-\sqrt{2},0)\cup(\sqrt{2},\infty)] and decreasing on [(-\infty,-\sqrt{2})\cup(0,\sqrt{2})].

Question 6.

Determine intervals where [f(x)=\sin x + \dfrac{x}{2}] is increasing (on [\mathbb{R}]).

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)][=\cos x + \dfrac{1}{2}].
2. Solve [\cos x + \dfrac{1}{2} > 0] ⇒ [\cos x > -\dfrac{1}{2}].
3. Cosine equals [ -\dfrac{1}{2}] at [x = \dfrac{2\pi}{3} + 2k\pi] and [x = \dfrac{4\pi}{3} + 2k\pi]. The inequality [\cos x > -\dfrac{1}{2}] holds between these roots modulo [2\pi].
4. Therefore [f(x)] is increasing on each union of intervals:
   [(-\dfrac{2\pi}{3} + 2k\pi,\ \dfrac{2\pi}{3} + 2k\pi)][\quad][\text{for every integer }k]
   (On the complementary arcs [\dfrac{2\pi}{3}+2k\pi,\dfrac{4\pi}{3}+2k\pi] the derivative is ≤0 so function is non-increasing there.)
5. Conclusion: [f(x)] is increasing on those intervals where [\cos x + \dfrac{1}{2} > 0] as written.

Question 7.

Show [f(x)=\arctan x] is strictly increasing on [\mathbb{R}].

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)][=\dfrac{1}{1+x^{2}}].
2. For all real [x], denominator [1+x^{2}>0] ⇒ derivative > 0.
3. Conclusion: Since derivative is positive everywhere, [f(x)] is strictly increasing on [(-\infty,\infty)].

Question 8.

Decide monotonicity of [f(x)=x^{3}] on [\mathbb{R}] and state whether it is strictly increasing.

Step-by-step Solution:

1. [\dfrac{d}{dx}f(x)=3x^{2}].
2. [3x^{2} \ge 0] for all x, and equals 0 only at [x=0].
3. Derivative is non-negative everywhere, but zero at an isolated point. To check strictness, take any [x_{2}>x_{1}]:
   [f(x_{2})-f(x_{1})][=(x_{2}-x_{1})(x_{2}^{2}+x_{1}x_{2}+x_{1}^{2})>0]
   because [x_{2}-x_{1}>0] and bracket is positive.
4. Conclusion: [f(x)=x^{3}] is strictly increasing on [(-\infty,\infty)] despite derivative being zero at one point.

Question 9.

Where is [f(x)][=\dfrac{x^{3}}{x^{2}+1}] increasing?

Step-by-step Solution:

1. Use quotient rule:
   [\dfrac{d}{dx}f(x)][=\dfrac{3x^{2}(x^{2}+1) – x^{3}(2x)}{(x^{2}+1)^{2}} ][= \dfrac{3x^{4}+3x^{2} -2x^{4}}{(x^{2}+1)^{2}} ][= \dfrac{x^{4}+3x^{2}}{(x^{2}+1)^{2}}].
2. Simplify numerator: [x^{2}(x^{2}+3)]. This is ≥ 0 for all x and equals 0 only at [x=0]. Denominator always > 0.
3. Therefore derivative ≥ 0 for all x and >0 for x≠0.
4. Since [f'(x)>0] for every x≠0 and only zero at the isolated point x=0, the function is strictly increasing on [(-\infty,0)]? wait — check sign: derivative nonnegative and zero only at 0, positive elsewhere, so function increases on entire real line (strictly).
5. Conclusion: [f(x)] is strictly increasing on [(-\infty,\infty)] (derivative positive except at isolated point).

Question 10.

Find where [f(x)=\dfrac{x}{1+x^{2}}] is increasing.

Step-by-step Solution:

1. Differentiate using quotient rule:
   [\dfrac{d}{dx}f(x)][=\dfrac{(1+x^{2})\cdot 1 – x\cdot 2x}{(1+x^{2})^{2}} ][= \dfrac{1+x^{2} -2x^{2}}{(1+x^{2})^{2}} ][= \dfrac{1 – x^{2}}{(1+x^{2})^{2}}].
2. Solve [1 – x^{2} > 0] ⇒ [|x|<1]. Derivative zero at [x=±1], negative for [|x|>1].
3. Conclusion: [f(x)] is increasing on [(-1,1)] (strictly increasing on that open interval) and decreasing on [(-\infty,-1)\cup(1,\infty)].